Find direction of initial velocity of nonrelativistic particle in B

[Quelsita]

I'm really trying to get my head around this one and our text doesn't give a very good example of this.

Problem:

A nonrelativistic particle of charge q and mass m is in a uniform magnetic field B in the z drection. The inital position is x=0, y=0, z=0; the magnitude of inital velocity of the particle is V₀.
What must be the direction of the initial velocity of the particle if it is to reach the point
x= a,2mv₀/qB, y=0, z=0 in the least possible time?


Assumtions:
1. If the transverse deflections are small, we can regard this velocity as constant and can say that the particle spends time t=l/v_z in the field.
2. The magnetic field gives the particle an acceleration in both the Y and Z direction which results in deflction in the Y direction?

The equation given in the text is tan(theta)_y= a_yt/v_x = eEl/mv²_x
but it honestly doesn't state what this is or how it is derived.

I know this is a lot of questioning, but any little bit helps. I'm still really trying to figure out what exactly is happening in the problem and how they came to use the equation.

Thanks.

 

[buffordboy23]

Assumtions:
1. If the transverse deflections are small, we can regard this velocity as constant and can say that the particle spends time t=l/v_z in the field.
2. The magnetic field gives the particle an acceleration in both the Y and Z direction which results in deflction in the Y direction?

According to the problem statement, the magnitude of the velocity is constant; it is the direction of the velocity vector that must be changing.

This vector equation describes the force on the particle:

$$\vec{F} = q\vec{v}\times\vec{B}$$

or equivalently, the magnitude of the force is

$$F = qvBsin\theta$$

Therefore, since the initial velocity is perpendicular to the magnetic field vector, there is no force in the z-direction. As a consequence, this reduces to a two-dimensional problem since the path of the particle is initially at z=0 and the path of the particle ends at z=0 for the z-dimension. So what you have is uniform circular motion in the xy-plane. Draw a picture to help see this; note that +q and -q revolve in different directions.

 

[gabbagabbahey]

The equation given in the text is tan(theta)_y= a_yt/v_x = eEl/mv²_x
but it honestly doesn't state what this is or how it is derived.

Thanks.

I see an E in this equation (perhaps one that is directed in the x direction?), is there also an electric field present here?

If so set up the equation of motion for the particle:

$$m \ddot{\vec{r}}=m (\ddot{x}\hat{x}+\ddot{y}\hat{y}+\ddot{z}\hat{z})= \vec{F}_{net}=q(\vec{E} + \vec{v} \times \vec{B})$$

Your given that $$\vec{B}=B\hat{z}$$, so

$$\vec{v} \times \vec{B}= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \dot{x} & \dot{y} & \dot{z} \\ 0 & 0 & B \end{vmatrix}=B\dot{y}\hat{x}-B\dot{x}\hat{y}$$

$$\Rightarrow m (\ddot{x}\hat{x}+\ddot{y}\hat{y}+\ddot{z}\hat{z})= q(\vec{E} +B\dot{y}\hat{x}-B\dot{x}\hat{y})$$

So just find the general solutions for $$x(t),y(t),z(t)$$ and substitute the initial conditions:

$$x(0)=y(0)=z(0)=0, \quad \dot{x}(0)=v_{0x}, \quad \dot{y}(0)=v_{0y}, \quad \dot{z}(0)=v_{0z}$$

to find the solutions for a given initial velocity $$v_0$$.

Then compute the time $$t_0$$ it takes to get to the point (a,0,0) and find which values of $$v_{0x},v_{0y},v_{0z}$$ minimize $$t_0$$.